Find dydx from first principles if y2x2Solutiondydx lim fxh

Find dy/dx from first principles if y=2x^2?

Solution

dy/dx = lim [f(x+h) - f(x)]/h, if h approaches to 0.

lim [f(x+h) - f(x)]/h = lim [2(x+h)^2 - 2x^2]/h

We\'ll expand the binomial:

lim [2(x+h)^2 - 2x^2]/h = lim (2x^2 + 4xh + 2h^2 - 2x^2)/h

We\'ll eliminate like terms inside brackets:

lim (2x^2 + 4xh + 2h^2 - 2x^2)/h = lim (4xh + 2h^2)/h

We\'ll factorize by 2h the numerator:

lim (4xh + 2h^2)/h = lim 2h*(2x + h)/h

We\'ll simplify and we\'ll get:

lim 2h*(2x + h)/h = lim 2*(2x + h)

We\'ll substitute h by the value of accumulation point:

lim 2*(2x + h) = 4x + 2*0

lim 2*(2x + h) = 4x

The value of dy/dx from first principles is: dy/dx = 4x.